Question

Two numbers are in the ratio 5:6. If 8 is subtracted from each of the numbers, the ratio becomes 4:5. Find the numbers.

Answer 1

Hint: In this question, we are given the ratios of two numbers. We are also given the ratio when 8 is subtracted from both the numbers. We need to find the actual numbers. For this, we will assume those numbers as x and y. After that, we will form the two equations using the given ratios. Solving these two equations will give us the values of x and y. We will use the substitution method for solving the equations. As we know that x:y can be written as \[\dfrac{x}{y},\] so we will use it to form the equations.

Complete step-by-step answer:
Here we are given the ratios of two numbers as 5:6. Let us assume that these two numbers are x and y. Hence, the ratio of x and y is given as 5:6.
\[\Rightarrow x:y::5:6\]
Now, we know that any ratio can be written in the fraction, i.e. x:y can be written as \[\dfrac{x}{y}.\]
So, x:y::5:6 can be written as \[\dfrac{x}{y}=\dfrac{5}{6}.\]
Cross multiplying, we get,
\[\Rightarrow 6x=5y\]
\[\Rightarrow 6x-5y=0.....\left( i \right)\]
This gives as an equation having variables x and y.
Now we are given that when 8 is subtracted from each of the numbers then the ratio is 4:5. So, when 8 is subtracted, the numbers become x – 8 and y – 8. And since the ratio of these numbers is 4:5. So,
\[x-8:y-8::4:5\]
Writing it in the form of the fraction, we get,
\[\Rightarrow \dfrac{x-8}{y-8}=\dfrac{4}{5}\]
Cross multiplying, we get,
\[\Rightarrow 5\left( x-8 \right)=4\left( y-8 \right)\]
\[\Rightarrow 5x-40=4y-32\]
Rearranging the terms, we get,
\[\Rightarrow 5x-4y=8.....\left( ii \right)\]
This gives us the second equation.
Now, let us solve the equations (i) and (ii) using the substituting method to get the values of x and y. From (i), we have
\[6x-5y=0\]
\[\Rightarrow 6x=5y\]
Dividing by 6 on both the sides, we get,
\[\Rightarrow x=\dfrac{5y}{6}\]
Putting this value of x in the equation (ii), we get,
\[\Rightarrow 5\left( \dfrac{5}{6}y \right)-4y=8\]
\[\Rightarrow \dfrac{25}{6}y-4y=8\]
Taking LCM of 6, we get,
\[\Rightarrow \dfrac{25y-24y}{6}=8\]
\[\Rightarrow \dfrac{y}{6}=8\]
Cross multiplying, we get,
\[\Rightarrow y=48\]
Putting in the value of y in equation (i), we get,
\[\Rightarrow 6x-5\left( 48 \right)=0\]
\[\Rightarrow 6x=240\]
\[\Rightarrow x=40\]
Hence the values of x and y are 40 and 48.
Therefore, the required numbers are 40 and 48.

Note: While forming equation (ii), make sure to subtract 8 from both the numbers. Students can also use the elimination method for solving this sum since equation (i) didn’t have any constant term, so the substitution method is easier than the elimination method. Students should take care of the signs while solving the equations. Take care while solving the fractional part with the integer part.